# Product of two independent Student distributions

What is the product of two independent student t distributions? In which case does this product product result in another t distribution?

• Because exact formulas are messy and there is no product that results in a t distribution, would you mind explaining the statistical motivation for this question? We might be able to suggest alternative approaches or more useful related questions.
– whuber
May 27 at 13:41
• Explaining why I need this product would be a little bit complicated, but I am looking for a special case when the product of 2 t distribution will give a t distribution or an unnormalized t distribution
– sam
May 27 at 15:18
• Could you clarify what an "unnormalized" t distribution might be?
– whuber
May 27 at 15:21
• I reformulated my problem so it will result in answering the following question: stats.stackexchange.com/questions/576853/…
– sam
May 27 at 16:37
• The distribution corresponding to the product of several $t$ densities is called a poly-t distribution. May 28 at 6:55

When $$X$$ and $$Y$$ are independent random variables with densities $$f_X$$ and $$f_Y,$$ the density of their product can be found with a change of variables as

$$f_{XY}(z) = \int_{\mathbb R} f_X(x) f_Y(z/x)\,\frac{\mathrm{d}x}{|x|}.$$

Ignoring normalizing constants (we'll consider these at the end), for two Student t densities with $$\nu$$ and $$\mu$$ degrees of freedom this integrand is proportional to

$$h(x,z) = \left(1 + \frac{x^2}{\mu}\right)^{-(\mu+1)/2}\, \left(1 + \frac{z^2}{x^2\nu}\right)^{-(\nu+1)/2}\,\frac{1}{|x|}.$$

Let's find a lower bound for $$f(z)$$ when $$z$$ is small. To do so, we may restrict the region of integration and we may replace the integrand by anything that never exceeds it.

Let $$z$$ be positive but less than $$1$$ and consider the integration region where $$x^2\nu$$ ranges between $$z^2$$ and $$1.$$ To get an appreciation for what's going on, here (with $$\mu=\nu=1$$) are plots of $$h(x,z)$$ for $$|z| = 1$$ (blue), $$1/2, 1/4,$$ and $$1/8$$ (red).

You can see that as $$|z|$$ approaches $$0,$$ there's more and more area pushed into this region. That's no surprise: we would expect the largest area (which corresponds to the highest density of the product) to be at the center of the product distribution, which (by symmetry) must be $$0.$$ But how large does it get?

Over the region $$x^2/\nu \in[z^2, 1]$$ the first factor of $$h$$ is smallest when $$x$$ is smallest and the second factor is smallest when $$x$$ is largest, whence throughout this region

\begin{aligned} h(x,z) &\ge \left(1 + \frac{z^2}{\mu\nu}\right)^{-(\mu+1)/2}\, \left(1 + \frac{z^2}{1}\right)^{-(\nu+1)/2}\,\frac{1}{|x|} \\ &\ge \left(1 + \frac{1}{\mu\nu}\right)^{-(\mu+1)/2}\, \left(1 + 1\right)^{-(\nu+1)/2}\,\frac{1}{|x|}. \end{aligned}

The second inequality is a consequence of $$z^2 \le 1.$$

The factors before $$1/|x|$$ are constant (but nonzero), depending only on $$\mu$$ and $$\nu,$$ so again let's consider them later and ignore them now. As $$x$$ varies over just the positive part of this region it runs from $$z/\sqrt{\nu}$$ to $$1/\sqrt{\nu},$$ giving a lower bound proportional to

$$\int_{z/\sqrt{\nu}}^{1/\sqrt{\nu}} \frac{\mathrm{d}x}{|x|} = -\log z.$$

As $$z\to 0,$$ this lower bound diverges. Consequently, no matter what the constants of proportionality are that we ignored, $$f_{XY}(z)$$ diverges at $$0.$$

Here, to illustrate, is a histogram from a simulation of ten million products (with $$\nu=\mu=1/2$$). Almost a million of those products are represented. The red curve is the negative logarithm. Clearly it approximates the density well near zero.

However, any Student t distribution with (say) $$\kappa \gt 0$$ degrees of freedom has a value proportional to $$(1 + 0^2/\kappa)^{-(\kappa+1)/2} = 1$$ at the origin, which is finite. Consequently, the product of two independent Student t distributions is never (even remotely like) a Student t distribution.

The product density can be found analytically as a polynomial combination of Riemann hypergeometric functions. Since this product is never a Student t distribution, though, I did not see any point into providing further details.

• Thank you for this detailed analysis. this makes sense now. For the special case when the degree of freedom is 1 therefore the t distribution is a cauchy distribution, it is possible to find the result of the product of two t(1)?
– sam
May 27 at 17:38
• @sam It seems so, see here: math.stackexchange.com/a/1451744 May 27 at 17:47
• Yes, that result is correct. But generally, for arbitrary degrees of freedom, the PDF of the product does not attain such a simple form: various versions of hypergeometric functions show up, such as elliptic functions.
– whuber
May 27 at 18:39